Find extrema of $f(x)=(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots+\frac{x^n}{n!})e^{-x}$

Question

$f(x)=(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots+\frac{x^n}{n!})e^{-x}$ where n ≥ 4 is a positive integer. Discuss the extremum of $f(x)$.

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My observations: Since $x$ is not very large we can't take it as $e^x$. If we had $n \to \infty$ then $f(x)$ would have been 1. How to proceed in this case where it not guaranteed that x is very large?

Answer 1

Let $g(n,x)=\sum_{i=0}^n \frac{x^i}{i!}$.
So $f(x)=e^{-x}\cdot g(n,x)$, and $$\frac{d}{dx}f(x)=g(x,n-1)\cdot e^{-x}-g(x,n-1)\cdot e^{-x}$$ $$=e^{-x}\bigl( g(n-1,x)-g(n,x)\bigl)=-\frac{e^{-x}\cdot x^n}{n!}$$ To find the extremum - $$e^{-x}\cdot x^n\overset!=0 \Rightarrow x=0$$

Answer 2

Hint By the first derivative test, it's sufficient to find and classify the critical points (the solutions of $f'(x) = 0$) of $$f(x) = T_n(x) e^{-x},$$ where $$T_n(x) := 1 + x + \frac{1}{2!} x^2 + \cdots + \frac{1}{n!} x^n. $$ Now, it's straightforward to compute that $T_n' = T_{n - 1}$ (for $n > 1$).

Additional hint It's also useful to observe that $T_n = T_{n - 1} + \frac{1}{n!} x^n$ (for $n \geq 1$).